Excel BI - Excel Challenge 822

excel-challenges

excel-formulas

🔰 Find the largest N digit numbers in square grids considering rows (left to right) and columns (top to bottom).

Published

March 24, 2026

Illustration for Excel BI - Excel Challenge 822

Challenge Description

🔰 Find the largest N digit numbers in square grids considering rows (left to right) and columns (top to bottom).

Solutions

library(tidyverse)
library(readxl)

get_list = . %>% unlist() %>% as.integer() %>% na.omit()

path = "Excel/800-899/822/822 Largest N Digit Numbers in Grids.xlsx"
input1 = read_excel(path, range = "A3:C5", col_names = FALSE) %>% as.matrix()
test1  = read_excel(path, range = "J3:M3", col_names = FALSE) %>% get_list()
input2 = read_excel(path, range = "A7:D10", col_names = FALSE) %>% as.matrix()
test2 = read_excel(path, range = "J7:M7", col_names = FALSE) %>% get_list()
input3 = read_excel(path, range = "A12:E16", col_names = FALSE) %>% as.matrix()
test3  = read_excel(path, range = "J12:M12", col_names = FALSE) %>% get_list()

roll_max_num <- function(v, n) {
  if (length(v) < n) return(-Inf)
  map_int(1:(length(v) - n + 1), ~ as.integer(paste0(v[.x:(.x + n - 1)], collapse = ""))) %>% max()
}

max_for_N <- function(mat, n) {
  rows <- asplit(mat, 1) %>% map_int(roll_max_num, n)
  cols <- asplit(mat, 2) %>% map_int(roll_max_num, n)
  max(c(rows, cols))
}

result1 = map_int(2:nrow(input1), ~ max_for_N(input1, .x))
result2 = map_int(2:nrow(input2), ~ max_for_N(input2, .x))
result3 = map_int(2:nrow(input3), ~ max_for_N(input3, .x))

all.equal(result1, test1, check.attributes = FALSE) # one different than provided
all.equal(result2, test2, check.attributes = FALSE) # all correct
all.equal(result3, test3, check.attributes = FALSE) # two different than provided

Logic: Read the workbook ranges needed for the challenge.
Strengths: The code maps the workbook rule into a compact, reproducible pipeline.
Areas for Improvement: The solution assumes the workbook layout and selected ranges remain stable, so any structural change in the sheet would require small adjustments.
Gem: The elegant part is how little code is needed once the correct intermediate representation is chosen.

import pandas as pd

path = "800-899/822/822 Largest N Digit Numbers in Grids.xlsx"
input1 = pd.read_excel(path, header=None, usecols="A:C", skiprows=2, nrows=3).values
test1  = pd.read_excel(path, header=None, usecols="J:M", skiprows=2, nrows=1).values.flatten(); test1 = [x for x in test1 if pd.notna(x)]
input2 = pd.read_excel(path, header=None, usecols="A:D", skiprows=6, nrows=4).values
test2  = pd.read_excel(path, header=None, usecols="J:M", skiprows=6, nrows=1).values.flatten(); test2 = [x for x in test2 if pd.notna(x)]
input3 = pd.read_excel(path, header=None, usecols="A:E", skiprows=11, nrows=5).values
test3 = pd.read_excel(path, header=None, usecols="J:M", skiprows=11, nrows=1).values.flatten(); test3 = [x for x in test3 if pd.notna(x)]

def roll_max_num(v, n):
    v = [str(x) for x in v]
    if len(v) < n:
        return float('-inf')
    nums = [int(''.join(v[i:i+n])) for i in range(len(v)-n+1)]
    return max(nums)

def max_for_N(mat, n):
    rows = [roll_max_num(row, n) for row in mat]
    cols = [roll_max_num(col, n) for col in mat.T]
    return max(rows + cols)

result1 = [max_for_N(input1, n) for n in range(2, input1.shape[0]+1)]
result2 = [max_for_N(input2, n) for n in range(2, input2.shape[0]+1)]
result3 = [max_for_N(input3, n) for n in range(2, input3.shape[0]+1)]

print(result1 == test1)  # one different than provided
print(result2 == test2)  # all correct
print(result3 == test3)  # two different than provided

The Python version keeps the algorithm explicit, which helps when the challenge depends on a greedy or iterative rule.

Difficulty Level

Easy / Medium

The business rule is clear, though the workbook still needs a few transformation steps to reach the expected output.